[Math] What does it mean when a system of linear equations have no solution

determinantlinear algebramatrices

$$
A =
\left(
\begin{array}{ccc}
10 & 29 & 41 \\
23 & 27 & 42 \\
24 & 28 & 48 \\
\end{array} \right)
$$
$\det (A) = -1748$.

Now $B$ is formed when the second column is multiplied by $15$ and added to the first column.
$$
B
= \left(
\begin{array}{ccc}
445 & 29 & 41 \\
428 & 27 & 42 \\
444 & 28 & 48 \\
\end{array} \right)
$$
$\det (B) = -1748$.

When i reduced these two matrices the final row is $[0,0,1]$, which means there are no solutions.

Why is that? If my system is inconsistent then what does it mean?

Best Answer

You are confusing reducing the matrix itself with reducing the augmented matrix --- the original matrix with the column of constants appended as the last column. If this augmented matrix reduces to a matrix with a row of zeros except in the last column, then the system is inconsistent and has no solutions.

In the case you've given, the matrix of coefficients $A$ has nonzero determinant, so you know (depending on what you've learned so far) that the system has a unique solution.

As far as what an inconsistent system means, consider the case of two equations in two unknowns. Each of those equations represents a line in the $xy$-plane, and a solution to the system is an intersection point of those lines. If the lines are nonparallel, they intersect in exactly one point and the system has a unique solution. If they coincide, there are an infinite number of solutions. But if they are parallel and not the same line, they do not intersect, so the system has no solutions and is inconsistent. The situation with more variables and more equations is more complicated when you try to visualize it, but is conceptually the same: if the surfaces defined have a unique intersection point, the system has a unique solution, and so on.